The Laplace transform of \( u(t) \) and its ROC is given as
Show Hint
- \( \frac{1}{s}, \, \sigma<0 \)
- \( \frac{1}{s}, \, \sigma>0 \)
- \( \frac{1}{s - 1}, \, \sigma = 0 \)
- \( \frac{1}{1 - s}, \, \sigma \leq 0 \)
The Correct Option is B
Solution and Explanation
To solve this problem, we need to analyze the Laplace transform of the step function \( u(t) \) and its Region of Convergence (ROC).
1. Understanding the Concepts:
- Laplace Transform of \( u(t) \): The step function \( u(t) \) is defined as:
\[ u(t) = \begin{cases} 0 & t < 0 \\ 1 & t \geq 0 \end{cases} \]
The Laplace transform of \( u(t) \) is given by:
\[ \mathcal{L}\{u(t)\} = \frac{1}{s} \]
- Region of Convergence (ROC): The ROC for the Laplace transform of a function is the range of values for which the Laplace transform converges. For \( u(t) \), the Laplace transform \( \frac{1}{s} \) converges when the real part of \( s \) is greater than 0, i.e., \( \sigma > 0 \), where \( s = \sigma + j\omega \) and \( \sigma \) is the real part.
2. Given Values:
The Laplace transform of \( u(t) \) is:
\[ \mathcal{L}\{u(t)\} = \frac{1}{s} \]
The ROC for this Laplace transform is \( \sigma > 0 \), as the integral converges for positive real parts of \( s \). This is because \( u(t) \) is a causal function that starts at \( t = 0 \) and its Laplace transform converges for \( \sigma > 0 \).
Final Answer:
The Laplace transform of \( u(t) \) and its ROC is \( \frac{1}{s}, \, \sigma > 0 \).
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